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Let $f(z)$ be a degree $3$ complex polynomial with real coefficients. How do we prove that $f(z)=0$ has three complex solutions without using the Fundamental Theorem of Algebra.

If $f(x)$ was real variable polynomial, I could use the Intermediate value theorem to at least get one root but with complex variable I have no idea what to do or how to start.

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  • $\begingroup$ You can solve the cubic by radicals using your favourite method - which is algebraically identical to the solution of a cubic with real coefficients $\endgroup$ – Mark Bennet Oct 28 '18 at 18:20
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Let $f(z)$ be a degree $3$ complex polynomial with real coefficients.

... tells you precisely that you can think of $f$ as a function $f\colon\mathbb R\to\mathbb R$ and can use intermediate value theorem.

After you obtain one real root $\alpha$, divide $f$ by $x-\alpha$ to get a degree $2$ polynomial. I assume you know how to prove how many roots it has.

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Note that real number field is a sub-field of complex numbers so it is OK to first search for real solutions.

Since your coefficients are real numbers you can find a real solution by looking at the end behavior of your polynomial and applying the intermediate value theorem.

Then by dividing your polynomial by $x- \alpha $ you find the other two solutions, which may be real or complex.

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